/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 57 Two circular loops of wire, each... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 21: Problem 57

Two circular loops of wire, each containing a single turn, have the same radius of \(4.0 \mathrm{cm}\) and a common center. The planes of the loops are perpendicular. Each carries a current of \(1.7 \mathrm{A}\). What is the magnitude of the net magnetic field at the common center?

Short Answer

Expert verified
The net magnetic field at the center is approximately \(3.77 \times 10^{-5} \text{ T}\).

Step by step solution

01

Understanding the Magnetic Field due to a Circular Loop

For a circular loop of wire, the magnetic field at the center of the loop is given by the formula: \[ B = \frac{\mu_0 I}{2R} \]where \(B\) is the magnetic field, \(\mu_0 = 4\pi \times 10^{-7} \) T路m/A (the permeability of free space), \(I\) is the current, and \(R\) is the radius of the loop.
02

Applying the Formula to Each Loop

Here, both loops have the same radius \(R = 4.0 \) cm (or \(0.04\) m) and current \(I = 1.7\) A. We calculate the magnetic field at the center due to one loop using:\[ B_1 = \frac{4\pi \times 10^{-7} \, \text{T} \, \text{m/A} \times 1.7 \, \text{A}}{2 \times 0.04 \, \text{m}} \approx 2.67 \times 10^{-5} \, \text{T} \]
03

Calculating the Net Magnetic Field

Since the loops are perpendicular, their magnetic fields are at right angles. We use the Pythagorean theorem to find the net magnetic field:\[ B_{net} = \sqrt{B_1^2 + B_2^2} = \sqrt{(2.67 \times 10^{-5} \text{ T})^2 + (2.67 \times 10^{-5} \text{ T})^2} \].
04

Solving for the Resulting Magnitude

By simplifying the expression, we find:\[ B_{net} = \sqrt{2 \times (2.67 \times 10^{-5} \text{ T})^2} = 2.67 \times \sqrt{2} \times 10^{-5} \, \text{T} \approx 3.77 \times 10^{-5} \, \text{T} \]. This is the magnitude of the net magnetic field at the center.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Circular Loop
A circular loop is essentially a wire bent into a perfect circle. In physics, particularly electromagnetism, circular loops are significant because they help create a magnetic field when an electric current flows through them. When current passes through the wire loop, it creates a magnetic field that is concentrated at the center of the loop.
This magnetic field can be calculated using the formula:
  • \( B = \frac{\mu_0 I}{2R} \), where:
    • \( B \) represents the magnetic field strength,
    • \( \mu_0 \) is the permeability of free space,
    • \( I \) is the current in amperes, and
    • \( R \) is the radius of the loop.
The relationship is straightforward; increasing the current or decreasing the radius will enhance the magnetic field strength. Circular loops are often used in devices like solenoids and electromagnets to efficiently produce controlled magnetic fields.
Permeability of Free Space
The permeability of free space, commonly denoted as \( \mu_0 \), is a physical constant that plays a critical role in the study of electromagnetism. It essentially describes how well a magnetic field can penetrate a vacuum, such as outer space.
In practical terms, permeability of free space helps us quantify the creation of a magnetic field in a vacuum. Its standard value is:
  • \( \mu_0 = 4\pi \times 10^{-7} \) T路m/A, where:
    • \( T \) is tesla, a unit measuring magnetic field strength,
    • \( m \) stands for meters, and
    • \( A \) is for amperes, unit of electric current.
The permeability of free space is essential in calculating the magnetic fields produced by currents. Alongside other constants, it contributes to equations that predict the behavior of magnetic phenomena in various settings.
Right Angle
A right angle is an angle of precisely 90 degrees, and it features prominently in physics and mathematics due to its unique properties.
In situations involving vectors, such as magnetic fields, right angles are crucial for calculations. When two vectors, representing magnetic fields, are at a right angle to each other, their combined effect is simpler to compute.
This is because forces or fields at right angles interact in a way that allows their magnitudes to be part of spatial dimensions, facilitating easier calculations due to their independence.
Pythagorean Theorem
The Pythagorean theorem is a fundamental principle in geometry that relates the sides of a right triangle. It states:
  • For a right triangle with legs of lengths \( a \) and \( b \), and hypotenuse \( c \), the relation holds: \( a^2 + b^2 = c^2 \).
In physics, this theorem is incredibly helpful in vector addition, especially when dealing with perpendicular vectors.
Consider our problem with two magnetic fields that exist at right angles due to the perpendicular loops. To find the net magnetic field at the center, we use the Pythagorean theorem:
  • \( B_{net} = \sqrt{B_1^2 + B_2^2} \)
This equation illustrates how combining two perpendicular magnetic fields results in a new field whose magnitude is easily computable using this ancient yet powerful theorem.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Electron beams are sometimes used to melt and evaporate metals in order to deposit thin metallic films on surfaces (similar to gold plating). One method is to put the material to be evaporated (called the "target") into a small tungsten cup (a crucible that has a very high melting point) and direct a beam of electrons at the target. Your team has been given the task of designing an electron-beam evaporator. The crucible is a cylinder, \(2.0 \mathrm{cm}\) in diameter and \(1.5 \mathrm{cm}\) in height, and contains a small target of pure nickel (Ni). The electrons are accelerated through a potential difference of \(V=1.20 \mathrm{kV}\), and form a beam that originates below the crucible, exactly \(3.70 \mathrm{cm}\) off its center, in the \(+x\) direction (see the drawing). (a) What is the speed of the electrons in the beam? (b) You must steer the electron beam with a magnetic field so that it curls over the lip of the cup and strikes the nickel target. Assuming that a uniform field exists above the cup (the field is zero below), what must be the radius of the beam's circular path? (c) In what direction should the field point if the beam initially approaches the cup from the \(-y\) axis? (d) What must be the magnitude of the uniform magnetic field?

Multiple-Concept Example 7 discusses how problems like this one can be solved. A \(+6.00 \mu \mathrm{C}\) charge is moving with a speed of \(7.50 \times 10^{4} \mathrm{m} / \mathrm{s}\) parallel to a very long, straight wire. The wire is \(5.00 \mathrm{cm}\) from the charge and carries a current of \(67.0 \mathrm{A}\) in a direction opposite to that of the moving charge. Find the magnitude and direction of the force on the charge.

Two circular coils are concentric and lie in the same plane. The inner coil contains 140 turns of wire, has a radius of \(0.015 \mathrm{m},\) and carries a current of \(7.2 \mathrm{A}\). The outer coil contains 180 turns and has a radius of \(0.023 \mathrm{m}\). What must be the magnitude and direction (relative to the current in the inner coil) of the current in the outer coil, so that the net magnetic field at the common center of the two coils is zero?

A horizontal wire of length \(0.53 \mathrm{m},\) carrying a current of \(7.5 \mathrm{A},\) is placed in a uniform external magnetic field. When the wire is horizontal, it experiences no magnetic force. When the wire is tilted upward at an angle of \(19^{\circ},\) it experiences a magnetic force of \(4.4 \times 10^{-3} \mathrm{N} .\) Determine the magnitude of the external magnetic field.

A charged particle enters a uniform magnetic field and follows the circular path shown in the drawing. (a) Is the particle positively or negatively charged? Why? (b) The particle's speed is \(140 \mathrm{m} / \mathrm{s},\) the magnitude of the magnetic field is \(0.48 \mathrm{T}\), and the radius of the path is \(960 \mathrm{m}\). Determine the mass of the particle, given that its charge has a magnitude of \(8.2 \times 10^{-4} \mathrm{C}\).
See all solutions

Recommended explanations on Physics Textbooks

Electrostatics

Read Explanation

Waves Physics

Read Explanation

Electricity and Magnetism

Read Explanation

Classical Mechanics

Read Explanation

Medical Physics

Read Explanation

Dynamics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.